limit $(x^2+y^2)e^{-(x+y)}$ when $(x,y)$ approach infinity

Question

Limit $(x^2+y^2)e^{-(x+y)}$ when $(x,y)$ approach infinity I don't know how to solve it. I've tried the following :

Set $x=r\cos\theta$ and $y=r\sin\theta$ and I've used $r$

In result, I got infinity.

Answer 1

Note that since $x,y > 0, x^2+y^2 \le(x+y)^2 \implies 0 < f(x,y) \le \dfrac{u^2}{e^{u}}$ by twice L’hospital rule with $u = x+y$ . Thus the limit is $0$ by Squeeze Theorem.

Answer 2

There is nothing wrong with the above solution. I tried to vote it up but I am a newbie ;). However it does not explain why you are wrong and often substituting bounds makes things more difficult conceptually.

1)The reason your answer is wrong is firstly, from the conditions you have given us, $x$ is not related to $y$. Under your subsititutions, $y^{2} = r^{2}-x^{2}$. Also it doesn't help you moveforward as the exponent is not in $x^{2}$ and $y^{2}$ but $x$ and $y$.

2) Your function can be written as $x^{2}e^{-x} \times y^{2}e^-{y}$. Both of these terms go to 0 as $x$ and $y$ go to infinity. The exponential dominates the polynomial. To see this intuitively you can go into $\log$ space, $\log(x^{2}) = 2 \log(x)$ and $\log(e^{-x}) = -x$, now you may be familiar with the $\log$ function, it tails off to grow less than linear quite quickly.